Clean link between electromotive force and difference of potential: potential difference around a resistor or inductance I would like to clarify some aspects of EM that I never realized before: proper link between voltage in electric circuit and electromotive force.
Electrical vision of electric circuit:
In almost all the sources, you see the following relationship in receptor convention:
The potential difference along an inductor is:
$$\Delta U=L \frac{d I}{d t}$$
The potential difference along a resistor:
$$\Delta U=R.I$$
EM vision of electrical circuits
Calling $A$ and $B$ the starting and ending point of a coil, we actually have:
$$e=\int_A^B \mathbf{E}.\mathbf{dl}=\oint \mathbf{E}.\mathbf{dl}=- \frac{\partial }{\partial t} \iint \mathbf{B} . \mathbf{dS} = - \frac{\partial \Phi}{\partial t}=-L \frac{dI}{dt}$$
Thus, I would not say $\Delta U=L \frac{d I}{d t}$, but rather: $e_L = -L \frac{dI}{dt}$, $e_L$ being the electromotive force of the inductance.
First question:
Is it a "mistake" when we see $\Delta U=L \frac{d I}{d t}$ with $\Delta U$ being the electric potential ? Shouldn't we only say $e_L = -L \frac{dI}{dt}$ ? Indeed, as we are not in electrostatics we don't have $e=-\Delta U$ here...

Now, for the resistor we can derive the macroscopic Ohm law from microscopic one. We have:
$$\mathbf{j}=\sigma \mathbf{E}$$
We assume $j$ uniform, thus $I=j*A$ for simplicity, with $A$ being the section of the conductor. We find:
$$e_R = \int_C^D \mathbf{E}.\mathbf{dl}  =\frac{1}{\sigma} \int_C^D  \mathbf{j} . \mathbf{dl}=\frac{L*I}{A*\sigma}=R*I$$
Where $R=\frac{L}{A*\sigma}$
Second question:
For the same reason, shouldn't we say, $e_R = R*I$ and not $\Delta U=R*I$ for a resistor ? The two expression will be equivalent in the static case but not in the electrodynamic one because of non conservation of the circulation of the electric field in this last case.
Note: I guess I miss a minus for the resistor because in the static case we have $e=-\Delta U$. I don't find the mistake at the moment but I guess it is an obvious one.
 A: In your calculations, if the loop is made of ideal wire, then $e=0$. 
You have to apply the path integral on an entire circuit, composed for example by an ideal inductor, a resistor, ideal wires, and a voltage generator.

Using Faraday's Law (as you can see, no electrostatics...), knowing that in wires (including inductor wires) the electric field is zero, one obtains:
$$\int_{\text{circuit}} \mathbf{E}\cdot \mathbf{l}_0 \mathrm{d}l = V_{\text{gen}}-RI = -\frac{\partial\Phi}{\partial t} =-\frac{\partial (L I)}{\partial t}$$
A: I think you're right; using $\Delta U$ when current isn't steady doesn't make sense. Every time I've seen Kirchoff's voltage law applied to AC circuits (for example, in Griffiths), it's been in terms of emf: $$\epsilon_0-RI-L\frac{dI}{dt}=0$$ I think the second and third terms would indeed be most correctly rewritten as $\epsilon_{R}$ and $\epsilon_{L}$.
To clear up why the TOTAL emf of the circuit is set equal to zero, besides saying "because Kirchoff said so": The magnetic flux through the circuit is negligible. The magnetic flux through the loops in the inductor, however, is obviously significant. This is why you get $\epsilon_{L}=-L\frac{dI}{dt}$. I hope this clears things up.
A: [edit]: I figured out that my explanation is consistent with:
from 5:45 to 13:45 : youtube.com/watch?v=t2micky_3uI, link suggested by @Ofek Gillon in the comment of the answer.
I will try to put a rigorous derivation of R.L circuit without using at all any results that comes from conservative fields. In practice I will never make use of the notion of electric potential which mainly makes sense for electrostatics because field is conservative. I will only work with electromotive forces.

I compute the electromotive force on the full circuit after having chosen an orientation as given in my image.
$$e_{tot}=\oint \mathbf{E}.\mathbf{dl} $$
The inductance is here to model the flux going to the full circuit. In presence of coils in the circuit, it will represent the flux going through those coils additionned to the flux going through the full circuit. Without coils in my circuit, it will represent the flux going through the full circuit. 
In all case we thus have:
$$e_{tot}=-L \frac{dI}{dt} $$
Now, I need to compute $\oint \mathbf{E}.\mathbf{dl}$.
$$e_{tot}=\oint \mathbf{E}.\mathbf{dl}=\int_{gen} \mathbf{E}.\mathbf{dl}+\int_R \mathbf{E}.\mathbf{dl}$$
For the resistor:
Here, I have no right to use $V_A-V_B = RI$ because I work in electrodynamic and the potential is not well defined electrically (field non conservative). Thus I need to "rederive" Ohm's law in term of electromotive forces. I assume the microscopic Ohm's law holds: $\mathbf{j}=\sigma \mathbf{E}$, and $\mathbf{j}$ uniform on a section of resistor.
$$\int_R \mathbf{E}.\mathbf{dl} = \int_A^B \mathbf{E}.\mathbf{dl} = \frac{1}{\sigma} \int_A^B \mathbf{j}.\mathbf{dl}=\frac{L I}{A \sigma} = R*I$$
Where $R=\frac{L}{A \sigma}$
Allright, the Ohm's law takes the same shape in electrodynamics.
Now for the generator:
$$\int_{gen} \mathbf{E}.\mathbf{dl}=-e_{gen}$$
Indeed, the electric field inside of the generator goes from + to - and is thus opposed to $\mathbf{dl}$.
In the end, I find:
$$-e_{gen}+R*I=-L \frac{dI}{dt} \Leftrightarrow e=R*I+L*\frac{dI}{dt}$$
Which is consistant with what we are supposed to have.
Remark:
Here we saw that the law for the resistor also holds in electrodynamic, it is the same for the capacitor, computing the e.f.m oriented from the +Q to -Q plate of a condensator we have as well $e_Q=\frac{Q}{C}$
A: In contrast to the other answers, it is actually perfectly legitimate to treat the voltage across an inductor in an AC circuit as a voltage instead of an EMF provided the assumptions of circuit theory are met. The assumptions of circuit theory are as follows (see Nilsson and Riedel, Electric Circuits, ch 1):
1) there is no net charge on any circuit element, 
2) there is no magnetic coupling between circuit elements, and 
3) the circuit is small compared to the speed of light and the time scales of interest
Those assumptions are very powerful and allow treating a circuit as a bunch of connected “lumped elements” with no internal structure and no geometry (only topology). 
One derivation is outlined in section 11.3 of the MIT online textbook http://web.mit.edu/6.013_book/www/book.html
There it shows how the circuit theory assumptions are used to simplify Maxwell’s equations such that at the boundary of the lumped element you can treat it simply by the voltage and current at a discrete number of terminals on the boundary with regard to electrical power. The magnetic fields inside the boundary are unimportant as are the voltages on the boundary anywhere else besides the terminals. 
So, although you may have an EMF inside the circuit element, as long as the assumptions of circuit theory are satisfied, it can be treated fully as a simple voltage anywhere outside the boundary of the lumped element, and in particular at the terminals. 
Note, Walter Lewin’s famous anti-Kirchoff’s lecture is not a contradiction to my comments here. His approach is to explicitly violate the assumptions of circuit theory. Contrary to his assertions, there is nothing wrong with Kirchoff’s circuit laws provided the assumptions of circuit theory are met. 
